1RB0RE 1LC1LD 0RA0LD 1LB0LA 1RF1RA ---1LB

Unsolved problem:

Does this TM halt? If so, how many steps does it take to halt?

1RB0RE_1LC1LD_0RA0LD_1LB0LA_1RF1RA_---1LB (bbch) is a probviously halting tetrational BB(6) Cryptid found by Racheline on 30 July 2024 (Discord link).

Analysis by Racheline

1RB0RE_1LC1LD_0RA0LD_1LB0LA_1RF1RA_---1LB
A(n,m) = 0^inf (01)^(3n-4) A> (01)^m 0^inf

A(2n,m) -> A(3n,m-2)
A(2n+1,m) -> A(3n+1,m-1)
A(n,0) -> A(2,3n-4)
A(n,-1) -> halt
start from A(2,0)

it's kinda like the original halting cryptid candidate - we can rewrite it like this:
a_0 = 2 and a_(i+1) = HydraMap(a_i)
b_0 = 0 and b_(i+1) = b_i+(1 if a_i is odd otherwise 2)
c_0 = 0 and c_(i+1) = 3a_j-4 where j is such that b_j = c_i

a: 2, 3, 4, 6, 9, 13, 19, 28, 42, 63, 94, 141, ...
b: 0, 2, 3, 5, 7,  8,  9, 10, 12, 14, 15,  17, ...
c: 0, 2, 5, 14, 185, 22205951667644132025548, ...

Code by N1vi

import time
import sys

sys.set_int_max_str_digits(1000000)

p=[4,7] #seq, replace with [2,0] for 1RB0RE_1LC1LD_0RA0LD_1LB0LA_1RF1RA_---1LB
s=-1 #step, do not change

while p[1]>=0:
	s+=1
	#print(p) #uncomment to allow printing
	#time.sleep(0.1) #uncomment to add time between printing
	if p[1]==0:
		p=[2,p[0]*3-4]
	elif p[0]%2:
		p[0]=p[0]//2*3+1
		p[1]-=1
	else:
		p[0]=p[0]//2*3
		p[1]-=2

#print(f'10^{len(str(p[0]))}') #approx
print(p[0]) #exact